Solubility of Ca5(PO4)3OH in g/100mL vs Ksp

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higherme
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The question is:

Calculate the solubility, in g/100mL, of Ca5(PO4)3OH.

when it asks for solubility, is that the same as asking for the Ksp of Ca5(PO4)3OH??

so basically, i have to find the concentration of Ca2+, PO4^3- and OH- and then find the Ksp ?

BUT, i thought Ksp have no units... how can it be in g/100mL then?

any one help please?
 
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What it is asking for is "how many grams of [itex]Ca_5(PO_4)_3OH[/itex] can you dissolve in 100 mL of water at standard room temperature?"
 
Try writing down the expression for Ksp as a first step.
 
Ca5(PO4)3OH <----> 5Ca2+ + 3PO4^3- + OH-

ksp = [Ca2+]^5 [PO4)3-]^3 [OH-]
ksp = (5X)^5 (3x)^3 (x)
6.80E-37 = 84375x^9
x = 2.72E-5 M = [Ca5(PO4)3OH ]

(2.72E-5 mol/L)* (502.307 g/mol) = 0.01366275g/L = 0.0136627g/1000mL

divide the top and bottom by 10 to get per 100mL:
0.0136627g/1000mL

1.37 E-3 g/100mL <---- my answer

am i doing it right? thanks